If 100 cm3 of oxygen pass through a porous plug in 50 seconds, the time taken for the same volume of hydrogen to pass through the same porous plug is? (O = ...

If 100 cm3 of oxygen pass through a porous plug in 50 seconds, the time taken for the same volume of hydrogen to pass through the same porous plug is?

(O = 16, H = 1)

Answer Details

The rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass. This means that lighter gases will effuse (or diffuse) faster than heavier gases under the same conditions. In this problem, we are given that 100 cm³ of oxygen pass through a porous plug in 50 seconds. We can use this information to determine the rate of effusion of oxygen. First, we need to convert the volume from cm³ to liters (L), since the units of gas rate are typically expressed in L/s or mL/s. 100 cm³ = 0.1 L Now we can calculate the rate of effusion of oxygen: Rate of effusion of oxygen = volume / time = 0.1 L / 50 s = 0.002 L/s Next, we need to use the rate of effusion of oxygen to determine the time it would take for the same volume of hydrogen to pass through the same porous plug. Since hydrogen has a lower molar mass than oxygen (1 g/mol versus 16 g/mol), we would expect it to effuse faster. In fact, the ratio of their effusion rates can be calculated using Graham's law of effusion: Rate of effusion of hydrogen / Rate of effusion of oxygen = sqrt (Molar mass of oxygen / Molar mass of hydrogen) Rate of effusion of hydrogen / 0.002 L/s = sqrt (16 g/mol / 1 g/mol) Rate of effusion of hydrogen / 0.002 L/s = 4 Rate of effusion of hydrogen = 4 x 0.002 L/s = 0.008 L/s Finally, we can use the rate of effusion of hydrogen to calculate the time it would take for the same volume (0.1 L) to pass through the porous plug: Time taken for hydrogen to pass through the porous plug = volume / rate of effusion of hydrogen Time taken for hydrogen to pass through the porous plug = 0.1 L / 0.008 L/s = 12.5 s Therefore, the answer is 12.5 s.